Calculate distance given 2 points, latitude and longitude [duplicate]

Question

Possible Duplicate:
MySQL latitude and Longitude table setup

I know this question has probably been asked many times, I've researched a lot and I need help with specific thing.

Lets say I have a form and user enters longitude and latitude, and I have a database which has table containing longitudes and latitudes, how would I search for a point or points in that table that are within 15 miles of radius?

Answer 1

You can use the formula that calculates distances between two points. For example:

function get_distance($latitude1, $longitude1, $latitude2, $longitude2, $unit = 'Mi') { 
    $theta = $longitude1 - $longitude2; 
    $distance = (sin(deg2rad($latitude1)) * sin(deg2rad($latitude2))) + 
                (cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * 
                cos(deg2rad($theta))); 
    $distance = acos($distance); 
    $distance = rad2deg($distance); 
    $distance = $distance * 60 * 1.1515; 
    switch($unit) { 
        case 'Mi': 
            break; 
        case 'Km' : 
            $distance = $distance * 1.609344; 
    } 
    return (round($distance,2)); 
}

You can also do something like:

$query = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) * 
            sin((`Latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * 
            cos((`Latitude`*pi()/180)) * cos(((".$longitude."- `Longitude`)* 
            pi()/180))))*180/pi())*60*1.1515
        ) as distance 
        FROM `MyTable` 
        HAVING distance >= ".$distance.";

Answer 2

If you have the Lat/Lon of two points, you can get delta Lat and delta Lon and convert this to a distance along latitude and a distance along longitude, and use Pythagorean theorem.

Have you looked at pages like http://www.movable-type.co.uk/scripts/latlong.html? This has several ways of computing distance. So as you go through your list of points, you use the formula to calculate the distance from each point to the point of interest and keep only those that satisfy R<15 miles.